1. Nov 28, 2012

### lomu123

This is a specific case involving a Jaguar, but any help on the correct way to find the radial axial and torque loads in a moving cars wheel would be greatly appreciated.

1. The problem statement, all variables and given/known data
Estimate the worst case radial, axial and torque loads on a wheel.

Mass of car : 1465 KG
Number of Wheels:4

2. Relevant equations
F=ma

3. The attempt at a solution

The radial load in each wheel is the results of the mass of the car. Since there are four wheel on a car when the car is static the load is evenly distributed between the four wheels. Since the mass of the Jaguar e-type is 1475 KG loaded it can be assumed the mass on each wheel is 369 KG. In order to calculate the force due to this mass we use Newton’s second law:

F=ma (1)

where a is acceleration due to gravity, 9.81 m/s2 , and m is the mass through each wheel, 369 KG.

F= 369 x 9.81

However in the case of our wheel we want to know the worst case radial load. This would be when the mass is not evenly distributed through the four wheels, but only 2. This would occur when the car is making a sharp turn causing one side to lift off the ground slightly. We can estimate at this point that the mass of the car is loaded onto 2 wheels giving a mass on each wheel of 737.5 KG. Using eq. 1 again we can estimate the radial load in each wheel:

F= 737.5 x 9.81

An axial load is produced when a car turns. This is due to the centripetal force produced when an object moves in a circular path. I shall estimate that the centripetal force is equal to that of the axial load in the wheels. In the case of the car this force is equal to that of the friction force produced by the tyre on the road. The equation for centripetal force is:

Fc = mac (2)

where ac = v2 /r giving:

Fc = (mv2)/r (3)

We can see from this equation that the maximum centripetal force (and therefore the maximum axial load) is when velocity is high and the turning radius, r, is small. In order to find suitable values for velocity and turning radius we have to equate the centripetal force to the frictional force. This gives:

μmg = (mv2)/r (4)

This can me re-arranged to give:

r = v2/μg (5)

Since we know the top speed of the car, 150 MPH (67 m/s) , and μ (the friction coefficient of a rubber tyre on dry tarmac is estimated to be 0.9) we can find the smallest turning radius of the car at that speed without slipping would be 508 m. If the car tried to turn a tighter corner than that at top speed it would slip and therefore my assumption of the centripetal force equalling the frictional force would not hold true. With this information we can now use eq. 3 to calculate the axial load of the car on the point of slipping, when it would be maximum.

Fc = (1475 x 67 x 67)/508 = 13034 N

We can assume the axial load is evenly distributed between the four wheels giving a maximum axial load of 3259N in each wheel.

The torque load is the wheel is given by the following equation:

T = Fr (6)

Where r is the radius of the wheel and F is given by eq.1 giving :

T = mar (7)

Since r and m are constant the maximum torque is present when the car accelerating or decelerating the most. In the case of a car this is going to be greatest when a car is braking. We can approximate the deceleration of a car at high speed using:

a = (v2 – u2) / 2s (8)

This assumes the deceleration is constant which in practice is not but shall be a suitable approximation in our case.
To find a appropriate value for s, the distance taken to stop, I referred to the national highway guidelines for breaking distances. For a car travelling 90 MPH (40.2 m/s) is takes a car on average 386m excluding reaction time. You could argue that the Jaguar produced in the 60’s would have the same quality brakes as a modern car but the Jaguar was built as a high performance sports car so I believe its brakes performance would be similar to that of a modern car. With this information we can calculate the deceleration:

a = -(40.2)2 / 2 x386 = -2.1 m/s

Since we know the mass of the car, 1475 KG, and the radius of the wheel, 0.1905m (excluding the rim of the wheel) we can use eq. 7 to give the maximum torque to be:
Tmax =1475 x 2.1 x 0.1905 = 590 N

2. Nov 28, 2012

### SteamKing

Staff Emeritus
1. Torque has units of force x distance.
2. You are assuming a 50-50 weight distribution front-rear.
3. You are neglecting the fact that on acceleration from stop, the engine torque to the driving wheels is multiplied through the gearbox and the differential on its way to the wheels.
4. Your calculation of braking torque in the OP assumes that all of this torque is applied to one wheel.